<!DOCTYPE html>
<html lang="en-US">
<!--********************************************-->
<!--*       Generated from PreTeXt source      *-->
<!--*                                          *-->
<!--*         https://pretextbook.org          *-->
<!--*                                          *-->
<!--********************************************-->
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<meta name="robots" content="noindex, nofollow">
</head>
<body class="ignore-math">
<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>(iii) Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation}
y^{\prime \prime}+b y^{\prime}+c y=e^{\alpha x} \cos \beta x ~P_n(x) \quad  (\textrm{discussions are similar for the case}~ e^{\alpha x} \sin \beta x~P_n(x) ).\tag{3.6.12}
\end{equation}
</div>
<p class="continuation">We note that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation*}
e^{i \beta x}=\cos \beta x+i \sin \beta x, \quad e^{-i \beta x}=\cos \beta x-i \sin \beta x \to \cos \beta x=\frac{e^{i \beta x}+e^{-i \beta x}}{2}.
\end{equation*}
</div>
<p class="continuation">Then</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation}
\begin{aligned}
&amp;y^{\prime \prime}+b y^{\prime}+c y=\frac{1}{2} e^{(\alpha+ i \beta) x} P_n(x) +\frac{1}{2} e^{(\alpha- i \beta) x} P_n(x) \quad (\textrm{can be decomposed into two parts})\\
&amp; \to y^{\prime \prime}+b y^{\prime}+c y=\frac{1}{2} e^{(\alpha+ i \beta) x} P_n(x)\\
&amp;\quad y^{\prime \prime}+b y^{\prime}+c y=\frac{1}{2} e^{(\alpha- i \beta) x} P_n(x) 
\end{aligned}\tag{3.6.13}
\end{equation}
</div>
<p class="continuation"><span class="process-math">\((\ref{eq3_22})_2\)</span> and <span class="process-math">\((\ref{eq3_22})_3\)</span> degenerate to case (ii).The discussions are as follows.<dfn class="terminology">If</dfn> <span class="process-math">\(\alpha+i \beta\)</span> is not a root of <span class="process-math">\(r^2+b r+c=0\text{,}\)</span> then correspondingly, <span class="process-math">\(\alpha-i \beta\)</span> is not a root to this equation neither. In this case, for <span class="process-math">\((\ref{eq3_22})_2\text{,}\)</span> the particular solution is given as</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation*}
y_1=e^{(\alpha+i \beta) x} (A_0 x^n+A_1 x^{n-1}+\cdots+A_{n-1} x+A_n),
\end{equation*}
</div>
<p class="continuation">and for <span class="process-math">\((\ref{eq3_22})_3\text{,}\)</span> the particular solution is given as</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation*}
y_2=e^{(\alpha-i \beta) x} (B_0 x^n+B_1 x^{n-1}+\cdots+B_{n-1} x+B_n),
\end{equation*}
</div>
<p class="continuation">Then the particular solution to (<a href="" class="xref" data-knowl="./knowl/eq3_23.html" title="Equation 3.6.12">(3.6.12)</a>) is given as</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation*}
\begin{aligned}
y=&amp;y_1+y_2\\
=&amp;\left[ A_0 e^{(\alpha+i \beta) x}+B_0 e^{(\alpha-i \beta) x}\right] x^n+\left[A_1e^{(\alpha+i \beta) x} +B_1 e^{(\alpha-i \beta) x}  \right] x^{n-1}+\cdots +\left[ A_n e^{(\alpha+i \beta) x}+B_n e^{(\alpha-i \beta) x}\right].
\end{aligned}
\end{equation*}
</div>
<p class="continuation">We may write</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation*}
\begin{aligned}
A_k e^{(\alpha+i \beta) x}+B_k e^{(\alpha-i \beta) x}&amp;= e^{\alpha x} \left[ (A_k+B_k) \cos \beta x+i (A_k-B_k) \sin \beta x \right]\\
&amp;=e^{\alpha x} \left[ C_k \cos \beta x+ D_k \sin \beta x  \right],
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(C_k=A_k+B_k\)</span> and <span class="process-math">\(D_k=i (A_k-B_k)\text{.}\)</span>Then the particular solution is rewritten as</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation*}
\begin{aligned}
y&amp;=e^{\alpha x} \left[ C_0 \cos \beta x+ D_0 \sin \beta x  \right] x^n+e^{\alpha x} \left[ C_1 \cos \beta x+ D_1 \sin \beta x  \right] x^{n-1}+\cdots+e^{\alpha x} \left[ C_n \cos \beta x+ D_n \sin \beta x  \right]\\
&amp;=e^{\alpha x} \cos \beta x \left[C_0 x^n+C_1 x^{n-1}+\cdots+C_n     \right]+e^{\alpha x} \sin \beta x \left[D_0 x^n+D_1 x^{n-1}+\cdots+D_n     \right].
\end{aligned}
\end{equation*}
</div>
<p class="continuation"><dfn class="terminology">If</dfn> <span class="process-math">\(\alpha + i\beta\)</span> is a root to <span class="process-math">\(r^2+b r+c=0\text{,}\)</span> then so is <span class="process-math">\(\alpha-i\beta\text{.}\)</span> In this case, we can get the particular solution is the form</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_23.html">
\begin{equation*}
y=e^{\alpha x} (\cos \beta x)~x \left[C_0 x^n+C_1 x^{n-1}+\cdots+C_n     \right]+e^{\alpha x} (\sin \beta x) ~x \left[D_0 x^n+D_1 x^{n-1}+\cdots+D_n     \right]
\end{equation*}
</div>
<span class="incontext"><a href="sec3_6.html#p-133" class="internal">in-context</a></span>
</body>
</html>
